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3 years ago

What the vertical height of a hemisphere

Mathematics

1 answer:

Sonja [21]3 years ago

3 0

Answer:

V= (2/3) π r³

hope this helps :>

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On a coordinate plane, the x-axis is labeled bags of trail mix and the y-axis is labeled ounces of almonds. Line a is labeled y

Musya8 [376]

Answer:

Line B

Step-by-step explanation:

Only one line on the graph shows 2 ounces of almonds for 1 bag of trail mix: line B.

7 0

3 years ago

Read 2 more answers

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

mote1985 [20]

Answer:

$\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}$

Step-by-step explanation:

To find the derivative of the function $y(x)=\ln \left(\frac{x}{x^2+1}\right)$ you must:

Step 1. Rewrite the logarithm:

$\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }$

Step 2. The derivative of a sum is the sum of derivatives:

$\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }}={\left(\left(\ln{\left(x \right)}\right)^{\prime } - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }\right)$

Step 3. The derivative of natural logarithm is $\left(\ln{\left(x \right)}\right)^{\prime }=\frac{1}{x}$

${\left(\ln{\left(x \right)}\right)^{\prime }} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }={\frac{1}{x}} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }$

Step 4. The function $\ln{\left(x^{2} + 1 \right)}$ is the composition $f\left(g\left(x\right)\right)$ of two functions $f\left(u\right)=\ln{\left(u \right)}$ and $u=g\left(x\right)=x^{2} + 1$

Step 5. Apply the chain rule $\left(f\left(g\left(x\right)\right)\right)^{\prime }=\frac{d}{du}\left(f\left(u\right)\right) \cdot \left(g\left(x\right)\right)^{\prime }$

$-{\left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }} + \frac{1}{x}=- {\frac{d}{du}\left(\ln{\left(u \right)}\right) \frac{d}{dx}\left(x^{2} + 1\right)} + \frac{1}{x}\\\\- {\frac{d}{du}\left(\ln{\left(u \right)}\right)} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- {\frac{1}{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}$

Return to the old variable:

$- \frac{1}{{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- \frac{\frac{d}{dx}\left(x^{2} + 1\right)}{{\left(x^{2} + 1\right)}} + \frac{1}{x}$

The derivative of a sum is the sum of derivatives:

$- \frac{{\frac{d}{dx}\left(x^{2} + 1\right)}}{x^{2} + 1} + \frac{1}{x}=- \frac{{\left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right)}}{x^{2} + 1} + \frac{1}{x}=\frac{1}{x^{3} + x} \left(x^{2} - x \left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right) + 1\right)$

Step 6. Apply the power rule $\frac{d}{dx}\left(x^{n}\right)=n\cdot x^{-1+n}$

$\frac{1}{x^{3} + x} \left(x^{2} - x \left({\frac{d}{dx}\left(x^{2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(x^{2} - x \left({\left(2 x^{-1 + 2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x \frac{d}{dx}\left(1\right) + 1\right)\\$

$\frac{1}{x^{3} + x} \left(- x^{2} - x {\frac{d}{dx}\left(1\right)} + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x {\left(0\right)} + 1\right)=\\\\\frac{1 - x^{2}}{x \left(x^{2} + 1\right)}$

Thus, $\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}$

3 0

3 years ago

the length of a rectangle exceeds its width by 6 inches and the area is 40 square inches. What are the length and width of the r

Lorico [155]

Classic Algebra and its unnecessarily complicated sentence structure. As you may have probably known, Algebra has its own "vocabulary set".

"the length of a rectangle exceeds its width by 6 inches" -> length is 6 in. longer than width -> l= w + 6

Since we're solving for the length and width, let's give them each variables.

length = l = w+6

width = w

The next bit of information is "the area is 40 square inches"

Applying the formula for the area of a rectangle we can set up:

l x w = 40

replace "l", or length, with it's alternate value.

(w+6) x w = 40

distribute

$w^{2}$ + 6w = 40

subtract 40 from both sides

$w^{2}$ + 6w - 40 = 0

factor

(w - 4)(w + 10) = 0

solve for w

w= 4, or -10

So great, we have 2 values; which one do we choose? Since this problem is referring to lengths and inches, we will have to choose the positive value. There is not such thing as a negative distance in the real world.

We now have half of the problem solved: width. Now we just need to find the length which we can do but substituting it back into the original alternate value of l.

l = w + 6

w=4

l = 4 + 6 = 10

The length is 10 in. and the width is 4 in. Hope this helps!

8 0

4 years ago

What is the measure of Angle A? (picture included)

Sergeu [11.5K]

Answer:

19.4

Step-by-step explanation:

10x-14=180

5 0

3 years ago

Read 2 more answers

Which equation is the inverse of y=100-x2

Harman [31]

Answer:

Inverse of y = $f^{-1}(x)$ = $\sqrt{100 -x}$ .

Step-by-step explanation:

Given : f(x) = y = 100 - x²

To find : Find inverse.

Solution : We have given that :

f(x) = y = 100 - x²

Step1) : Replace every x with a y and replace every y with an x .

x = 100 - y².

Step 2): Solve for y

On subtracting 100 from both sides

x - 100 = -y²

On multiplying both side by -1

100 -x = y²

Taking square root both sides and switching sides

y = ± $\sqrt{100-x}$ .

Step3 ) : Replace y with $f^{-1}(x)$ .

$f^{-1}(x)$ = ± $\sqrt{100-x}$ .

Therefore, inverse of y = $f^{-1}(x)$ = ± $\sqrt{100-x}$ .

3 0

4 years ago